Question

Find local maximum and local minimum values of the function given by \[ f(x) = 3x^4 + 4x^3 - 12x^2 + 12 \]

Hint:

To find local maxima and minima, use the first derivative to find critical points, and the second derivative to determine whether each point is a maximum or minimum.

Answer 1

Step 1: Find the first derivative of the function.
To find the local maxima and minima, we first find the first derivative of the function \( f(x) \): \[ f'(x) = \frac{d}{dx}(3x^4 + 4x^3 - 12x^2 + 12) = 12x^3 + 12x^2 - 24x \]

Step 2: Set the first derivative equal to zero.
To find the critical points, we solve \( f'(x) = 0 \): \[ 12x^3 + 12x^2 - 24x = 0 \] Factor out \( 12x \): \[ 12x(x^2 + x - 2) = 0 \] Now, solve the quadratic equation \( x^2 + x - 2 = 0 \): \[ (x - 1)(x + 2) = 0 \] Thus, \( x = 0, 1, -2 \).

Step 3: Use the second derivative to determine the nature of the critical points.
The second derivative of \( f(x) \) is: \[ f''(x) = \frac{d}{dx}(12x^3 + 12x^2 - 24x) = 36x^2 + 24x - 24 \]

Step 4: Check the values of \( f''(x) \) at the critical points.
- At \( x = 0 \): \[ f''(0) = 36(0)^2 + 24(0) - 24 = -24 \text{(local maximum)}. \] - At \( x = 1 \): \[ f''(1) = 36(1)^2 + 24(1) - 24 = 36 \text{(local minimum)}. \] - At \( x = -2 \): \[ f''(-2) = 36(-2)^2 + 24(-2) - 24 = 144 - 48 - 24 = 72 \text{(local minimum)}. \]

Step 5: Conclusion.
- Local maximum occurs at \( x = 0 \) with \( f(0) = 12 \). - Local minima occur at \( x = 1 \) with \( f(1) = 3 \) and at \( x = -2 \) with \( f(-2) = 48 \).